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An ion having charge +6e is traveling horizontally to the left at 8.05 km/s when it...

An ion having charge +6e is traveling horizontally to the left at 8.05 km/s when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of 6.84×10−15 N . What is the magnitude of this field?

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Answer #1

use,
v= 8.05 km/s
= 8050 m/s
F = q*v*B*sin theta
where,
theta is angle between velocity and field
since, given
velocity is in left direction and field is directed downward
so,
theta = 90

so, by using above formula
6.84×10^-15 = 6*1.6*10^-19* 8050*B
6.84×10^-15 = 7.73*10^-15*B
B = 0.885 tesla

Answer: 0.885 tesla

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